Marcin 'Qrczak' Kowalczyk wrote:
>
> What a pattern matches is independent from which of the variables
> it binds are used.
>
> The pattern [a,b,c], or equivalently a::b::c::[], matches lists of
> length 3.
> (...)
... unless it matches lazily, as in
let [a,b,c] = [1,2,3,4] in let [a
Hi,
I think a lazy pattern is "really lazy" if reference to a part of the
pattern does not
cause matching of rest of the pattern.
For example,
\ ~((:) a ((:) b [])) -> [a]
should not try to match ((:) b[]) part of the pattern, since 'a' is only
referenced.
It seems the above pattern should
Patrik Jansson writes:
>No, I did not! The program is tested in hugs and ghci and does what it
>"should".
You are right; I was wrong.
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On Sat, 20 Oct 2001, Richard wrote:
> Patrik Jansson writes:
> >This also works: (the (mis)match with the empty list is delayed
> > until the end of time)
> >
> >> test5= let a: b: ~(c: ~[]) = "Hello"
> >>in [a,b,c]
> ^^^
> typo! You meant [a,b].
No, I d
I just wrote:
>Patrik Jansson writes:
>
>>> test5= let a: b: ~(c: ~[]) = "Hello"
>>>in [a,b,c]
> ^^^
>typo! You meant [a,b].
Never mind! Now I see the ~ in front of the [].
Must look more closely before posting.
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Patrik Jansson writes:
>This also works: (the (mis)match with the empty list is delayed
> until the end of time)
>
>> test5= let a: b: ~(c: ~[]) = "Hello"
>>in [a,b,c]
^^^
typo! You meant [a,b].
>But this fails (if the full list is requested ...)
>
>> t
On Sat, 20 Oct 2001, Adrian Hey wrote:
...
: Interesting. I always thought pattern matching was lazy, but it
I don't think Haskell has changed here (recently).
> main = print test3
This does not work:
> test1= let [a,b,c] = "Hello"
>in [a,b]
Same here:
> test2= let a: b: c: [] = "Hel
On Saturday 20 October 2001 3:58 am, Saswat Anand wrote:
> Hi,
> I am wondering why this function should not work with input [x,y]
> (list with two elements) too, since third element is not referenced. Why
> is it so eager to pattern match.
>
> fun = \list -> let [a,b,c] = list
>
Sat, 20 Oct 2001 10:58:02 +0800 (GMT-8), Saswat Anand <[EMAIL PROTECTED]> pisze:
> I am wondering why this function should not work with input [x,y]
> (list with two elements) too, since third element is not referenced.
What a pattern matches is independent from which of the variables
it binds
Hi,
I am wondering why this function should not work with input [x,y]
(list with two elements) too, since third element is not referenced. Why
is it so eager to pattern match.
fun = \list -> let [a,b,c] = list
in [a,b]
Thanks,
Saswat
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